Why is this an NFA, but not an FA? The 2019 Stack Overflow Developer Survey Results Are InConvert from DFA to NFAProve that a PDA with accept states accepts all context-free languagesHow can a DFA corresponding to an NFA have a transition that the original NFA does not?Intersection of 2 deterministic finite state automata, but nondeterministicallyFinite automata as dynamical systemsLanguages acceptable with just a single final stateFind equivalence classes of given regular expressionWhy is this transition considered to be non deterministic?Converting NFA to DFA. Loops and exits.Given a transition table do digraph, determine if it is DFA or NFA and build grammar
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Why is this an NFA, but not an FA?
The 2019 Stack Overflow Developer Survey Results Are InConvert from DFA to NFAProve that a PDA with accept states accepts all context-free languagesHow can a DFA corresponding to an NFA have a transition that the original NFA does not?Intersection of 2 deterministic finite state automata, but nondeterministicallyFinite automata as dynamical systemsLanguages acceptable with just a single final stateFind equivalence classes of given regular expressionWhy is this transition considered to be non deterministic?Converting NFA to DFA. Loops and exits.Given a transition table do digraph, determine if it is DFA or NFA and build grammar
$begingroup$
It doesn't have non-deterministic paths.
Every transition is one-to-one.
Accepted states are defined/guaranteed on specific inputs.
Not sure why its not an FA
Edit: 0.0.2
automata
asked Mar 30 at 21:59
A_for_ AbacusA_for_ Abacus
9671025
$endgroup$
add a comment |
$begingroup$
It doesn't have non-deterministic paths.
Every transition is one-to-one.
Accepted states are defined/guaranteed on specific inputs.
Not sure why its not an FA
Edit: 0.0.2
automata
asked Mar 30 at 21:59
A_for_ AbacusA_for_ Abacus
9671025
$endgroup$
2
$begingroup$
There are no transitions from the accepting state.
$endgroup$
– saulspatz
Mar 30 at 22:23
$begingroup$
@saul I think the convention is that any missing transitions are taken as going to a "black hole" state, a non-accepting state that can't be left.
$endgroup$
– Gerry Myerson
Mar 30 at 23:09
$begingroup$
It might help if we could see 0.0.2
$endgroup$
– Gerry Myerson
Mar 30 at 23:10
$begingroup$
@GerryMyerson I added 0.0.2, still don't understand why its not a FA.
$endgroup$
– A_for_ Abacus
Mar 30 at 23:28
add a comment |
$begingroup$
It doesn't have non-deterministic paths.
Every transition is one-to-one.
Accepted states are defined/guaranteed on specific inputs.
Not sure why its not an FA
Edit: 0.0.2
automata
asked Mar 30 at 21:59
A_for_ AbacusA_for_ Abacus
9671025
$endgroup$
It doesn't have non-deterministic paths.
Every transition is one-to-one.
Accepted states are defined/guaranteed on specific inputs.
Not sure why its not an FA
Edit: 0.0.2
automata
automata
asked Mar 30 at 21:59
A_for_ AbacusA_for_ Abacus
9671025
asked Mar 30 at 21:59
A_for_ AbacusA_for_ Abacus
9671025
edited Mar 30 at 23:15
A_for_ Abacus
asked Mar 30 at 21:59
A_for_ AbacusA_for_ Abacus
9671025
asked Mar 30 at 21:59
A_for_ AbacusA_for_ Abacus
9671025
asked Mar 30 at 21:59
A_for_ AbacusA_for_ Abacus
9671025
9671025
2
$begingroup$
There are no transitions from the accepting state.
$endgroup$
– saulspatz
Mar 30 at 22:23
$begingroup$
@saul I think the convention is that any missing transitions are taken as going to a "black hole" state, a non-accepting state that can't be left.
$endgroup$
– Gerry Myerson
Mar 30 at 23:09
$begingroup$
It might help if we could see 0.0.2
$endgroup$
– Gerry Myerson
Mar 30 at 23:10
$begingroup$
@GerryMyerson I added 0.0.2, still don't understand why its not a FA.
$endgroup$
– A_for_ Abacus
Mar 30 at 23:28
add a comment |
2
$begingroup$
There are no transitions from the accepting state.
$endgroup$
– saulspatz
Mar 30 at 22:23
$begingroup$
@saul I think the convention is that any missing transitions are taken as going to a "black hole" state, a non-accepting state that can't be left.
$endgroup$
– Gerry Myerson
Mar 30 at 23:09
$begingroup$
It might help if we could see 0.0.2
$endgroup$
– Gerry Myerson
Mar 30 at 23:10
$begingroup$
@GerryMyerson I added 0.0.2, still don't understand why its not a FA.
$endgroup$
– A_for_ Abacus
Mar 30 at 23:28
2
2
$begingroup$
There are no transitions from the accepting state.
$endgroup$
– saulspatz
Mar 30 at 22:23
$begingroup$
There are no transitions from the accepting state.
$endgroup$
– saulspatz
Mar 30 at 22:23
$begingroup$
@saul I think the convention is that any missing transitions are taken as going to a "black hole" state, a non-accepting state that can't be left.
$endgroup$
– Gerry Myerson
Mar 30 at 23:09
$begingroup$
@saul I think the convention is that any missing transitions are taken as going to a "black hole" state, a non-accepting state that can't be left.
$endgroup$
– Gerry Myerson
Mar 30 at 23:09
$begingroup$
It might help if we could see 0.0.2
$endgroup$
– Gerry Myerson
Mar 30 at 23:10
$begingroup$
It might help if we could see 0.0.2
$endgroup$
– Gerry Myerson
Mar 30 at 23:10
$begingroup$
@GerryMyerson I added 0.0.2, still don't understand why its not a FA.
$endgroup$
– A_for_ Abacus
Mar 30 at 23:28
$begingroup$
@GerryMyerson I added 0.0.2, still don't understand why its not a FA.
$endgroup$
– A_for_ Abacus
Mar 30 at 23:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Nvm, figured it out.
The transition function $delta$ must be total in a FA.
In the 1st example, at the accepting state $q_1$, $delta(q_1, 0)=undefined$.
Whereas in the 2nd example, if we label the states $q_0, q_1,q_2$ from left to right, at the accepting state $q_1$, $delta(q_1, 0)=q_2$.
answered Mar 30 at 23:40
A_for_ AbacusA_for_ Abacus
9671025
$endgroup$
1
$begingroup$
OK. Not every author uses that convention. As I wrote in the comments, some authors use the convention that missing transitions go to a state like the one you call $q_2$ and I call a "black hole".
$endgroup$
– Gerry Myerson
Mar 31 at 2:26
2
$begingroup$
I'm going off on a tangent, and it's a pretty trivial thing anyway, but it's never correct to write "$= undefined$", because "undefined" isn't an expression or a variable, it's an adjective. It's a bit like writing "$6 = even$".
$endgroup$
– Tanner Swett
Mar 31 at 3:59
add a comment |
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$begingroup$
Nvm, figured it out.
The transition function $delta$ must be total in a FA.
In the 1st example, at the accepting state $q_1$, $delta(q_1, 0)=undefined$.
Whereas in the 2nd example, if we label the states $q_0, q_1,q_2$ from left to right, at the accepting state $q_1$, $delta(q_1, 0)=q_2$.
answered Mar 30 at 23:40
A_for_ AbacusA_for_ Abacus
9671025
$endgroup$
1
$begingroup$
OK. Not every author uses that convention. As I wrote in the comments, some authors use the convention that missing transitions go to a state like the one you call $q_2$ and I call a "black hole".
$endgroup$
– Gerry Myerson
Mar 31 at 2:26
2
$begingroup$
I'm going off on a tangent, and it's a pretty trivial thing anyway, but it's never correct to write "$= undefined$", because "undefined" isn't an expression or a variable, it's an adjective. It's a bit like writing "$6 = even$".
$endgroup$
– Tanner Swett
Mar 31 at 3:59
add a comment |
$begingroup$
Nvm, figured it out.
The transition function $delta$ must be total in a FA.
In the 1st example, at the accepting state $q_1$, $delta(q_1, 0)=undefined$.
Whereas in the 2nd example, if we label the states $q_0, q_1,q_2$ from left to right, at the accepting state $q_1$, $delta(q_1, 0)=q_2$.
answered Mar 30 at 23:40
A_for_ AbacusA_for_ Abacus
9671025
$endgroup$
1
$begingroup$
OK. Not every author uses that convention. As I wrote in the comments, some authors use the convention that missing transitions go to a state like the one you call $q_2$ and I call a "black hole".
$endgroup$
– Gerry Myerson
Mar 31 at 2:26
2
$begingroup$
I'm going off on a tangent, and it's a pretty trivial thing anyway, but it's never correct to write "$= undefined$", because "undefined" isn't an expression or a variable, it's an adjective. It's a bit like writing "$6 = even$".
$endgroup$
– Tanner Swett
Mar 31 at 3:59
add a comment |
$begingroup$
Nvm, figured it out.
The transition function $delta$ must be total in a FA.
In the 1st example, at the accepting state $q_1$, $delta(q_1, 0)=undefined$.
Whereas in the 2nd example, if we label the states $q_0, q_1,q_2$ from left to right, at the accepting state $q_1$, $delta(q_1, 0)=q_2$.
answered Mar 30 at 23:40
A_for_ AbacusA_for_ Abacus
9671025
$endgroup$
Nvm, figured it out.
The transition function $delta$ must be total in a FA.
In the 1st example, at the accepting state $q_1$, $delta(q_1, 0)=undefined$.
Whereas in the 2nd example, if we label the states $q_0, q_1,q_2$ from left to right, at the accepting state $q_1$, $delta(q_1, 0)=q_2$.
answered Mar 30 at 23:40
A_for_ AbacusA_for_ Abacus
9671025
answered Mar 30 at 23:40
A_for_ AbacusA_for_ Abacus
9671025
answered Mar 30 at 23:40
A_for_ AbacusA_for_ Abacus
9671025
answered Mar 30 at 23:40
A_for_ AbacusA_for_ Abacus
9671025
9671025
1
$begingroup$
OK. Not every author uses that convention. As I wrote in the comments, some authors use the convention that missing transitions go to a state like the one you call $q_2$ and I call a "black hole".
$endgroup$
– Gerry Myerson
Mar 31 at 2:26
2
$begingroup$
I'm going off on a tangent, and it's a pretty trivial thing anyway, but it's never correct to write "$= undefined$", because "undefined" isn't an expression or a variable, it's an adjective. It's a bit like writing "$6 = even$".
$endgroup$
– Tanner Swett
Mar 31 at 3:59
add a comment |
1
$begingroup$
OK. Not every author uses that convention. As I wrote in the comments, some authors use the convention that missing transitions go to a state like the one you call $q_2$ and I call a "black hole".
$endgroup$
– Gerry Myerson
Mar 31 at 2:26
2
$begingroup$
I'm going off on a tangent, and it's a pretty trivial thing anyway, but it's never correct to write "$= undefined$", because "undefined" isn't an expression or a variable, it's an adjective. It's a bit like writing "$6 = even$".
$endgroup$
– Tanner Swett
Mar 31 at 3:59
1
1
$begingroup$
OK. Not every author uses that convention. As I wrote in the comments, some authors use the convention that missing transitions go to a state like the one you call $q_2$ and I call a "black hole".
$endgroup$
– Gerry Myerson
Mar 31 at 2:26
$begingroup$
OK. Not every author uses that convention. As I wrote in the comments, some authors use the convention that missing transitions go to a state like the one you call $q_2$ and I call a "black hole".
$endgroup$
– Gerry Myerson
Mar 31 at 2:26
2
2
$begingroup$
I'm going off on a tangent, and it's a pretty trivial thing anyway, but it's never correct to write "$= undefined$", because "undefined" isn't an expression or a variable, it's an adjective. It's a bit like writing "$6 = even$".
$endgroup$
– Tanner Swett
Mar 31 at 3:59
$begingroup$
I'm going off on a tangent, and it's a pretty trivial thing anyway, but it's never correct to write "$= undefined$", because "undefined" isn't an expression or a variable, it's an adjective. It's a bit like writing "$6 = even$".
$endgroup$
– Tanner Swett
Mar 31 at 3:59
add a comment |
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2
$begingroup$
There are no transitions from the accepting state.
$endgroup$
– saulspatz
Mar 30 at 22:23
$begingroup$
@saul I think the convention is that any missing transitions are taken as going to a "black hole" state, a non-accepting state that can't be left.
$endgroup$
– Gerry Myerson
Mar 30 at 23:09
$begingroup$
It might help if we could see 0.0.2
$endgroup$
– Gerry Myerson
Mar 30 at 23:10
$begingroup$
@GerryMyerson I added 0.0.2, still don't understand why its not a FA.
$endgroup$
– A_for_ Abacus
Mar 30 at 23:28